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受欢迎的三角函数 >

cot^2(θ)+4csc(θ)=-5

解答

cot^{2} (θ) + 4 csc (θ) = - 5

解答

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

+1

度数

θ = 21 0^{\circ} + 36 0^{\circ} n, θ = 33 0^{\circ} + 36 0^{\circ} n

求解步骤

cot^{2} (θ) + 4 csc (θ) = - 5

两边减去

- 5

cot^{2} (θ) + 4 csc (θ) + 5 = 0

使用三角恒等式改写

5 + cot^{2} (θ) + 4 csc (θ)

使用毕达哥拉斯恒等式:

1 + cot^{2} (x) = csc^{2} (x)

cot^{2} (x) = csc^{2} (x) - 1

= 5 + csc^{2} (θ) - 1 + 4 csc (θ)

化简

5 + csc^{2} (θ) - 1 + 4 csc (θ) : csc^{2} (θ) + 4 csc (θ) + 4

5 + csc^{2} (θ) - 1 + 4 csc (θ)

对同类项分组

= csc^{2} (θ) + 4 csc (θ) + 5 - 1

数字相加/相减:

5 - 1 = 4

= csc^{2} (θ) + 4 csc (θ) + 4

= csc^{2} (θ) + 4 csc (θ) + 4

4 + csc^{2} (θ) + 4 csc (θ) = 0

用替代法求解

4 + csc^{2} (θ) + 4 csc (θ) = 0

令

: csc (θ) = u

4 + u^{2} + 4 u = 0

4 + u^{2} + 4 u = 0 : u = - 2

4 + u^{2} + 4 u = 0

改写成标准形式

a x^{2} + b x + c = 0

u^{2} + 4 u + 4 = 0

使用求根公式求解

u^{2} + 4 u + 4 = 0

二次方程求根公式:

若

a = 1, b = 4, c = 4

u_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot 4}{2 \cdot 1}

u_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot 4}{2 \cdot 1}

4^{2} - 4 \cdot 1 \cdot 4 = 0

4^{2} - 4 \cdot 1 \cdot 4

数字相乘:

4 \cdot 1 \cdot 4 = 16

= 4^{2} - 16

4^{2} = 16

= 16 - 16

数字相减:

16 - 16 = 0

= 0

u_{1, 2} = \frac{- 4 \pm 0}{2 \cdot 1}

u = \frac{- 4}{2 \cdot 1}

\frac{- 4}{2 \cdot 1} = - 2

\frac{- 4}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{- 4}{2}

使用分式法则:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{4}{2}

数字相除:

\frac{4}{2} = 2

= - 2

u = - 2

二次方程组的解是:

u = - 2

u = csc (θ)

代回

csc (θ) = - 2

csc (θ) = - 2

csc (θ) = - 2 : θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

csc (θ) = - 2

csc (θ) = - 2

的通解

csc (x)

周期表（周期为

2 πn

）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} csc (x) Undefiend 22 \frac{2 3}{3} 1 \frac{2 3}{3} 22 x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} csc (x) Undefiend - 2 - 2 - \frac{2 3}{3} - 1 - \frac{2 3}{3} - 2 - 2

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

合并所有解

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

作图

流行的例子

10sin(a)+7sin(a)+1sin(a)+8sin(a)=48

10 sin (a) + 7 sin (a) + 1 sin (a) + 8 sin (a) = 48

2cos^2(θ)+7sin(θ)=5

2 cos^{2} (θ) + 7 sin (θ) = 5

(2sqrt(3))/3 =sec(θ)

\frac{2 3}{3} = sec (θ)

4(sin(x))^2-15sin(x)-4=0

4 (sin (x))^{2} - 15 sin (x) - 4 = 0

1+sin(θ)=1+sin(-θ)

1 + sin (θ) = 1 + sin (- θ)